from __future__ import print_function, division
from mpmath.libmp import (fzero,
from_man_exp, from_int, from_rational,
fone, fhalf, bitcount, to_int, to_str, mpf_mul, mpf_div, mpf_sub,
mpf_add, mpf_sqrt, mpf_pi, mpf_cosh_sinh, pi_fixed, mpf_cos,
mpf_sin)
from sympy.core.numbers import igcd
from sympy.core.compatibility import range
from .residue_ntheory import (_sqrt_mod_prime_power,
legendre_symbol, jacobi_symbol, is_quad_residue)
import math
def _pre():
maxn = 10**5
global _factor
global _totient
_factor = [0]*maxn
_totient = [1]*maxn
lim = int(maxn**0.5) + 5
for i in range(2, lim):
if _factor[i] == 0:
for j in range(i*i, maxn, i):
if _factor[j] == 0:
_factor[j] = i
for i in range(2, maxn):
if _factor[i] == 0:
_factor[i] = i
_totient[i] = i-1
continue
x = _factor[i]
y = i//x
if y % x == 0:
_totient[i] = _totient[y]*x
else:
_totient[i] = _totient[y]*(x - 1)
def _a(n, k, prec):
""" Compute the inner sum in HRR formula [1]_
References
==========
.. [1] http://msp.org/pjm/1956/6-1/pjm-v6-n1-p18-p.pdf
"""
if k == 1:
return fone
k1 = k
e = 0
p = _factor[k]
while k1 % p == 0:
k1 //= p
e += 1
k2 = k//k1 # k2 = p^e
v = 1 - 24*n
pi = mpf_pi(prec)
if k1 == 1:
# k = p^e
if p == 2:
mod = 8*k
v = mod + v % mod
v = (v*pow(9, k - 1, mod)) % mod
m = _sqrt_mod_prime_power(v, 2, e + 3)[0]
arg = mpf_div(mpf_mul(
from_int(4*m), pi, prec), from_int(mod), prec)
return mpf_mul(mpf_mul(
from_int((-1)**e*jacobi_symbol(m - 1, m)),
mpf_sqrt(from_int(k), prec), prec),
mpf_sin(arg, prec), prec)
if p == 3:
mod = 3*k
v = mod + v % mod
if e > 1:
v = (v*pow(64, k//3 - 1, mod)) % mod
m = _sqrt_mod_prime_power(v, 3, e + 1)[0]
arg = mpf_div(mpf_mul(from_int(4*m), pi, prec),
from_int(mod), prec)
return mpf_mul(mpf_mul(
from_int(2*(-1)**(e + 1)*legendre_symbol(m, 3)),
mpf_sqrt(from_int(k//3), prec), prec),
mpf_sin(arg, prec), prec)
v = k + v % k
if v % p == 0:
if e == 1:
return mpf_mul(
from_int(jacobi_symbol(3, k)),
mpf_sqrt(from_int(k), prec), prec)
return fzero
if not is_quad_residue(v, p):
return fzero
_phi = p**(e - 1)*(p - 1)
v = (v*pow(576, _phi - 1, k))
m = _sqrt_mod_prime_power(v, p, e)[0]
arg = mpf_div(
mpf_mul(from_int(4*m), pi, prec),
from_int(k), prec)
return mpf_mul(mpf_mul(
from_int(2*jacobi_symbol(3, k)),
mpf_sqrt(from_int(k), prec), prec),
mpf_cos(arg, prec), prec)
if p != 2 or e >= 3:
d1, d2 = igcd(k1, 24), igcd(k2, 24)
e = 24//(d1*d2)
n1 = ((d2*e*n + (k2**2 - 1)//d1)*
pow(e*k2*k2*d2, _totient[k1] - 1, k1)) % k1
n2 = ((d1*e*n + (k1**2 - 1)//d2)*
pow(e*k1*k1*d1, _totient[k2] - 1, k2)) % k2
return mpf_mul(_a(n1, k1, prec), _a(n2, k2, prec), prec)
if e == 2:
n1 = ((8*n + 5)*pow(128, _totient[k1] - 1, k1)) % k1
n2 = (4 + ((n - 2 - (k1**2 - 1)//8)*(k1**2)) % 4) % 4
return mpf_mul(mpf_mul(
from_int(-1),
_a(n1, k1, prec), prec),
_a(n2, k2, prec))
n1 = ((8*n + 1)*pow(32, _totient[k1] - 1, k1)) % k1
n2 = (2 + (n - (k1**2 - 1)//8) % 2) % 2
return mpf_mul(_a(n1, k1, prec), _a(n2, k2, prec), prec)
def _d(n, j, prec, sq23pi, sqrt8):
"""
Compute the sinh term in the outer sum of the HRR formula.
The constants sqrt(2/3*pi) and sqrt(8) must be precomputed.
"""
j = from_int(j)
pi = mpf_pi(prec)
a = mpf_div(sq23pi, j, prec)
b = mpf_sub(from_int(n), from_rational(1, 24, prec), prec)
c = mpf_sqrt(b, prec)
ch, sh = mpf_cosh_sinh(mpf_mul(a, c), prec)
D = mpf_div(
mpf_sqrt(j, prec),
mpf_mul(mpf_mul(sqrt8, b), pi), prec)
E = mpf_sub(mpf_mul(a, ch), mpf_div(sh, c, prec), prec)
return mpf_mul(D, E)
[docs]def npartitions(n, verbose=False):
"""
Calculate the partition function P(n), i.e. the number of ways that
n can be written as a sum of positive integers.
P(n) is computed using the Hardy-Ramanujan-Rademacher formula [1]_.
The correctness of this implementation has been tested through 10**10.
Examples
========
>>> from sympy.ntheory import npartitions
>>> npartitions(25)
1958
References
==========
.. [1] http://mathworld.wolfram.com/PartitionFunctionP.html
"""
n = int(n)
if n < 0:
return 0
if n <= 5:
return [1, 1, 2, 3, 5, 7][n]
if '_factor' not in globals():
_pre()
# Estimate number of bits in p(n). This formula could be tidied
pbits = int((
math.pi*(2*n/3.)**0.5 -
math.log(4*n))/math.log(10) + 1) * \
math.log(10, 2)
prec = p = int(pbits*1.1 + 100)
s = fzero
M = max(6, int(0.24*n**0.5 + 4))
if M > 10**5:
raise ValueError("Input too big") # Corresponds to n > 1.7e11
sq23pi = mpf_mul(mpf_sqrt(from_rational(2, 3, p), p), mpf_pi(p), p)
sqrt8 = mpf_sqrt(from_int(8), p)
for q in range(1, M):
a = _a(n, q, p)
d = _d(n, q, p, sq23pi, sqrt8)
s = mpf_add(s, mpf_mul(a, d), prec)
if verbose:
print("step", q, "of", M, to_str(a, 10), to_str(d, 10))
# On average, the terms decrease rapidly in magnitude.
# Dynamically reducing the precision greatly improves
# performance.
p = bitcount(abs(to_int(d))) + 50
return int(to_int(mpf_add(s, fhalf, prec)))
__all__ = ['npartitions']
